Решебник по алгебре 8 класс Макарычев ФГОС Задание 67

 

\[\boxed{\text{67\ (67).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[Тождество:\]

\[\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c};\]

\[\frac{a - b}{c} = \frac{a}{c} - \frac{b}{c}.\]

Решение.

\[\textbf{а)}\ \frac{x^{2} + y^{2}}{x^{4}} = \frac{x^{2}}{x^{4}} + \frac{y^{2}}{x^{4}} = \frac{1}{x^{2}} + \frac{y^{2}}{x^{4}}\]

\[\textbf{б)}\ \frac{2x - y}{b} = \frac{2x}{b} - \frac{y}{b}\]

\[\textbf{в)}\ \frac{a^{2} + 1}{2a} = \frac{a^{2}}{2a} + \frac{1}{2a} = \frac{a}{2} + \frac{1}{2a}\]

\[\textbf{г)}\ \frac{a^{2} - 3ab}{a^{3}} = \frac{a(a - 3b)}{a^{3}} =\]

\[= \frac{a - 3b}{a^{2}} = \frac{a}{a^{2}} - \frac{3b}{a^{2}} = \frac{1}{a} - \frac{3b}{a^{2}}\]

\[\boxed{\text{67.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{x^{2}}{x^{2} - 16} - \frac{8 \cdot (x - 2)}{x^{2} - 16} =\]

\[= \frac{x^{2} - 8 \cdot (x - 2)}{x^{2} - 16} =\]

\[= \frac{x^{2} - 8x + 16}{x^{2} - 16} =\]

\[\textbf{б)}\ \frac{64 - 2ab}{(a - 8)^{2}} + \frac{2ab - a^{2}}{(8 - a)^{2}} =\]

\[= \frac{64 - 2ab}{(a - 8)^{2}} + \frac{2ab - a^{2}}{(a - 8)^{2}} =\]

\[= \frac{64 - 2ab + 2ab - a^{2}}{(a - 8)^{2}} =\]

\[= \frac{8 + a}{8 - a}\]